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A group of 8 friends want to play doubles tennis. How many

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A group of 8 friends want to play doubles tennis. How many [#permalink] New post 04 Jun 2004, 12:36
A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105


This doesnt seem very hard, but the final piece has me wondering.

1st Team: 8C2 = 28 (pick 2 from the 8)
2nd Team: 6C2 = 15 (pick 2 from the remaining 6)
3rd Team: 4C2 = 6 (pick 2 from the remaining 4)
4th Team: 2C2 = 1 (pick 2 from the remaining 2)

Multiply them together = 28*15*6*1 = 2520.

The answer is E: 105.

Can someone please explain why the answer is not simply 2520? I read the explanation but it doesnt really explain the concept of why the answer is derived as such.
Thanks.
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 [#permalink] New post 04 Jun 2004, 13:00
4 groups are indistinguishable--that is, all 16 permutations of <G1, G2, G3, G4> should be counted as one.

Hence 2520/4! = 105.

Another Q from gmattutor:
THere are 6 profs, divided into three groups, with exactly two in each group. Each group selects exactly one of the following subjects to teach - SCIENCE, HISTORY & LITERATURE.

In this case, the groups are distinguishable. Hence the ans is 6c2*4c2*2c2; that is, <G1, G2, G3> and <G2, G3, G1> are different, for G1 in the former teach Science, whereas G1 in the latter teach Literature.
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 [#permalink] New post 04 Jun 2004, 13:52
Thanks Hallelujah. I'm still a bit fuzzy on the definition of "distinguishable."

I assume that, in the tennis example, each group contains tennis players (non distinguishable) vs. in the teacher example each group contains different subjects (distinguishable).

However, shouldnt the tennis elements be classified as distinguishable? AKA - there are different people (e.g. you, me, Agassi, Kournikova, etc) playing....thus no two groups are the same?
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 [#permalink] New post 04 Jun 2004, 13:58
You can toy with a simple problem--say 4 people, divide them into 2 groups of 2 people. Of course, it is fuzzy, unless you solve it in brute force method.
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 [#permalink] New post 04 Jun 2004, 20:47
Another way of looking at the solution to the problem: -

The number of ways in which m+n+p+q things can be divided into 4 groups containing m,n,p,q things severally is (m+n+p+q)!/(m!*n!*p!)

When m=n=p=q, then the formula takes the form (4m)!/(m!*m!*m!*m!)
But this formula regards as different all the possible orders in which the 4 groups can occur in any one mode of sub-division. And since there are 4! such orders corresponding to each sub-division, the number of different ways in which subdivision into 4 equal groups can be made is

(4m)!/(m!*m!*m!*m!*4!)

So, applying this to the problem we have m = 2
and (4*2)!/(2!*2!*2!*2!*4!) = 40320/384 = 105
_________________

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  [#permalink] 04 Jun 2004, 20:47
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